Points A, B and C are on circle with centre O such that $$\angle$$BOC = 84$$^\circ$$. If AC is produced to a point D such that $$\angle$$BDC = 40$$^\circ$$, then find the measure of $$\angle$$ABD (in degrees).

Angle subtended by chord BC at the centre is twice the angle subtended by chord BC on the point A of the circle.

$$\angle$$BOC = 2$$\angle$$BAC

84$$^\circ$$ = 2$$\angle$$BAC

$$\angle$$BAC = 42$$^\circ$$

From triangle BAD,

$$\angle$$BAD + $$\angle$$ABD + $$\angle$$BDA = 180$$^\circ$$

42$$^\circ$$ + $$\angle$$ABD + 40$$^\circ$$ = 180$$^\circ$$

$$\angle$$ABD = 98$$^\circ$$

Hence, the correct answer is Option A